Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main(int argc, char *argv[]) {
  ios::sync_with_stdio(0);
  int n, m;
  cin >> n >> m;
  int mat[101][101], mat1[101][101];
  vector<pair<int, int> > mov, mov1;
  for (int i = 0; (i) < (n); ++i)
    for (int j = 0; (j) < (m); ++j) {
      cin >> mat[i][j];
      mat1[i][j] = mat[i][j];
    }
  for (int i = 0; (i) < (n); ++i) {
    int mn = INT_MAX;
    for (int j = 0; (j) < (m); ++j) {
      mn = min(mn, mat[i][j]);
    }
    if (mn < INT_MAX && mn > 0) {
      for (int j = 0; (j) < (m); ++j) mat[i][j] -= mn;
      for (int j = 0; (j) < (mn); ++j) mov.push_back(make_pair(0, i + 1));
    }
  }
  for (int i = 0; (i) < (m); ++i) {
    int mn = INT_MAX;
    for (int j = 0; (j) < (n); ++j) {
      mn = min(mn, mat[j][i]);
    }
    if (mn < INT_MAX && mn > 0) {
      for (int j = 0; (j) < (n); ++j) mat[j][i] -= mn;
      for (int j = 0; (j) < (mn); ++j) mov.push_back(make_pair(1, i + 1));
    }
  }
  for (int i = 0; (i) < (m); ++i) {
    int mn = INT_MAX;
    for (int j = 0; (j) < (n); ++j) {
      mn = min(mn, mat1[j][i]);
    }
    if (mn < INT_MAX && mn > 0) {
      for (int j = 0; (j) < (n); ++j) mat1[j][i] -= mn;
      for (int j = 0; (j) < (mn); ++j) mov1.push_back(make_pair(1, i + 1));
    }
  }
  for (int i = 0; (i) < (n); ++i) {
    int mn = INT_MAX;
    for (int j = 0; (j) < (m); ++j) {
      mn = min(mn, mat1[i][j]);
    }
    if (mn < INT_MAX && mn > 0) {
      for (int j = 0; (j) < (m); ++j) mat1[i][j] -= mn;
      for (int j = 0; (j) < (mn); ++j) mov1.push_back(make_pair(0, i + 1));
    }
  }
  bool possible = true;
  for (int i = 0; (i) < (n); ++i)
    for (int j = 0; (j) < (m); ++j)
      if (mat[i][j] != 0) possible = false;
  bool possible1 = true;
  for (int i = 0; (i) < (n); ++i)
    for (int j = 0; (j) < (m); ++j)
      if (mat1[i][j] != 0) possible1 = false;
  if (possible && !possible1) {
    cout << ((int)(mov).size()) << endl;
    for (int i = 0; (i) < (((int)(mov).size())); ++i)
      cout << ((!mov[i].first) ? "row" : "col") << " " << mov[i].second << endl;
  } else if (possible1 && !possible) {
    cout << ((int)(mov1).size()) << endl;
    for (int i = 0; (i) < (((int)(mov1).size())); ++i)
      cout << ((!mov1[i].first) ? "row" : "col") << " " << mov1[i].second
           << endl;
  } else if (possible && possible1) {
    if (((int)(mov1).size()) < ((int)(mov).size())) {
      cout << ((int)(mov1).size()) << endl;
      for (int i = 0; (i) < (((int)(mov1).size())); ++i)
        cout << ((!mov1[i].first) ? "row" : "col") << " " << mov1[i].second
             << endl;
    } else {
      cout << ((int)(mov).size()) << endl;
      for (int i = 0; (i) < (((int)(mov).size())); ++i)
        cout << ((!mov[i].first) ? "row" : "col") << " " << mov[i].second
             << endl;
    }
  } else
    cout << "-1" << endl;
  return 0;
}